1. ## solving equation

am i using a valid alternative solution?

question:

$\displaystyle \sin x+\cos\frac{x}{2}=0$

correct solution:

$\displaystyle \sin(2\cdot\frac{x}{2})+cos\frac{x}{2}=0$

$\displaystyle \2sin\frac{x}{2}\cos\frac{x}{2}+cos\frac{x}{2}=0$

$\displaystyle \cos\frac{x}{2}(\2sin\frac{x}{2}+1)=0$

$\displaystyle \cos\frac{x}{2}=0, \sin\frac{x}{2}=-\frac{1}{2}$

since $\displaystyle 0\leq x<2\pi, \sin\frac{x}{2}\geq 0$

$\displaystyle \sin\frac{x}{2}\neq -\frac{1}{2}$

$\displaystyle x=\pi$

my solution (is this correct?):

$\displaystyle \sin x+\sin(\frac{\pi}{2}\cdot\frac{x}{2})=0$

$\displaystyle 2\sin\frac{x+\pi}{4}=0, \cos\frac{3x-\pi}{4}=0$

$\displaystyle \frac{\pi}{4}\leq\frac{x+\pi}{4}<\frac{3}{4}\pi, /frac{\pi}{4}\leq\frac{3x-\pi}{4}<\frac{5}{4}\pi$

$\displaystyle \frac{x+\pi}{4}=0, \frac{3x-\pi}{4}=\frac{\pi}{2}$

since $\displaystyle 0\leq x<2\pi, x\neq -\pi$

$\displaystyle x=\pi$

2. ## Re: solving equation

Originally Posted by muddywaters
my solution (is this correct?):
$\displaystyle \sin x+\sin(\frac{\pi}{2}\cdot\frac{x}{2})=0$
What have you done here?

An altenative solution could be to use the fact that $\displaystyle \cos\left(\frac{x}{2}\right)=\sin\left(\frac{\pi}{ 2}-\frac{x}{2}\right)$

3. ## Re: solving equation

Originally Posted by Siron
What have you done here?

An altenative solution could be to use the fact that $\displaystyle \cos\left(\frac{x}{2}\right)=\sin\left(\frac{\pi}{ 2}-\frac{x}{2}\right)$
i'm sorry you're right. typo >< it should be $\displaystyle \cos\left(\frac{x}{2}\right)=\sin\left(\frac{\pi}{ 2}-\frac{x}{2}\right)$

4. ## Re: solving equation

with that, is it a valid alternative solutioN?

5. ## Re: solving equation

ish. just realized the mistakes in the first post. it should me liek this:

my solution (is this correct?):

$\displaystyle \sin x+\sin(\frac{\pi}{2}-\frac{x}{2})=0$

$\displaystyle 2\sin\frac{x+\pi}{4}\cos\frac{3x-\pi}{4}=0$

$\displaystyle 2\sin\frac{x+\pi}{4}=0, \cos\frac{3x-\pi}{4}=0$

$\displaystyle \frac{\pi}{4}\leq\frac{x+\pi}{4}<\frac{3}{4}\pi, \frac{\pi}{4}\leq\frac{3x-\pi}{4}<\frac{5}{4}\pi$

$\displaystyle \frac{x+\pi}{4}=0, \frac{3x-\pi}{4}=\frac{\pi}{2}$

since $\displaystyle 0\leq x<2\pi, x\neq -\pi$

$\displaystyle x=\pi$

6. ## Re: solving equation

Your solution looks fine to me.